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%I #29 Oct 31 2022 14:00:36
%S 1,1,1,1,2,2,1,6,12,12,1,24,144,288,288,1,120,2880,17280,34560,34560,
%T 1,720,86400,2073600,12441600,24883200,24883200,1,5040,3628800,
%U 435456000,10450944000,62705664000,125411328000,125411328000
%N Triangle read by rows: T(n,k) = Product_{i=n-k+1..n} i! for 0 <= k <= n.
%C Based on some integer sequence a(n), n>0, define triangular arrays A(a;n,k) by recurrence: A(a;0,0) = 1, and A(a;i,j) = 0 if j<0 or j>i, and A(a;n,k) = n! / (n-k)! * A(a;n-1,k) + a(n) * A(a;n-1,k-1) for 0<=k<=n. Then, Product_{i=1..n} (1 + (a(i) / i!) * x) = Sum_{k=0..n} A(a;n,k) / T(n,k) * x^k for n>=0 with empty product 1 (case n=0).
%C For the row reversed triangle R(n,k) = Product_{i=k+1..n} i! with empty product 1 (case k=n) the terms of the matrix inverse M are given by M(n,n) = 1 for n >= 0 and M(n,n-1) = -n! for n > 0 otherwise 0. - _Werner Schulte_, Oct 25 2022
%F T(n,k) = T(n,1) * T(n-1,k-1) for 0 < k <= n.
%F T(2*n,n) = A093002(n+1) for n >= 0.
%F T(n,k)/T(k,k) = A009963(n,k) for 0 <= k <= n.
%F (Sum_{k=0..n} T(n,k) * T(n,n-k))/T(n,n) = A193520(n) for n >= 0.
%e The triangle starts:
%e n\k : 0 1 2 3 4 5 6
%e ============================================================
%e 0 : 1
%e 1 : 1 1
%e 2 : 1 2 2
%e 3 : 1 6 12 12
%e 4 : 1 24 144 288 288
%e 5 : 1 120 2880 17280 34560 34560
%e 6 : 1 720 86400 2073600 12441600 24883200 24883200
%e etc.
%t T[n_, k_] := Product[i!, {i, n - k + 1, n}]; Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* _Amiram Eldar_, Jul 08 2020 *)
%Y Cf. A000012 (col_0), A000142 (col_1), A010790 (col_2), A176037 (col_3), A000178 (main diagonal and first subdiagonal).
%Y Row sums equal A051399(n+1).
%Y Cf. A009963, A093002, A193520.
%K nonn,easy,tabl
%O 0,5
%A _Werner Schulte_, Jul 08 2020